Question

A cannon of mass m1 = 12000 kg fires a shell of mass m2=300 kg with a velocity of V2=400m/s.Find the velocity of the cannon after it is shot​

Answer 1

Step-by-step explanation:

By law of conservation of momentum ,

m1u1 + m2u2 = m1v1 + m2v2 (INITIAL MOMENTUM = FINAL MOMENTUM)

Here,m1 = mass of cannon = 12000kg

m2 = mass of shell = 300kg

u1 = initial velocity of cannon i.e. 0 because cannon is initially at rest

u2 = initial velocity of shell i.e. 0 because initially shell is also at rest

v1 = Recoil velocity of cannon = ?

v2 = Final velocity of shell after it is shot = 400m/s

Put the above given values in the equation of law of conservation of momentum

=>12000×0 + 300×0=12000×v1 + 300×400

=>0+0=12000v1 + 120000

=> - 120000 = 12000v1

=> v1 = -10 m/s ans.

v1 is negative because cannon recoils(moves) in the direction opposite to that of shell.

Answer 2

✔️Answer ✔️

Your answer:-

Check out the attachment please.

We know that

Initial Momentum=Final Momentum

<< 0=m1v1-m2v2

<< m1v1=m2v2

<< v1=m2v2/m1

<< 300 x 400/1200

After cutting

v1=10ms^-1

For Extra Knowledge:-

ꜰᴇᴡ ɪɴᴅɪᴠɪᴅᴜᴀʟꜱ ʜᴀᴠᴇ ʜᴀᴅ ᴀꜱ ᴘʀᴏꜰᴏᴜɴᴅ ᴀɴ ɪᴍᴘᴀᴄᴛ ᴏɴ ꜱᴄɪᴇɴᴄᴇ ᴀꜱ ɢᴀʟɪʟᴇᴏ, ᴡʜᴏꜱᴇ ɢʀᴏᴜɴᴅʙʀᴇᴀᴋɪɴɢ ɪɴᴠᴇɴᴛɪᴏɴꜱ ᴀɴᴅ ᴅɪꜱᴄᴏᴠᴇʀɪᴇꜱ ᴇᴀʀɴᴇᴅ ʜɪᴍ ᴛʜᴇ ᴛɪᴛʟᴇ 'ᴛʜᴇ ꜰᴀᴛʜᴇʀ ᴏꜰ ꜱᴄɪᴇɴᴄᴇ'. ɢᴀʟɪʟᴇᴏ ᴡᴀꜱ ᴀɴ ᴇxᴘᴇʀɪᴍᴇɴᴛᴀʟɪꜱᴛ ᴡʜᴏ ꜰᴏʀ ᴛʜᴇ ꜰɪʀꜱᴛ ᴛɪᴍᴇ ʜᴀᴅ ᴛʜᴇ ɪɴꜱɪɢʜᴛ ᴀɴᴅ ᴛᴀʟᴇɴᴛ ᴛᴏ ʟɪɴᴋ ᴛʜᴇᴏʀʏ ᴡɪᴛʜ ᴇxᴘᴇʀɪᴍᴇɴᴛ.

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